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Introduction to Biostatistics The 2nd Lecture

HiroyoshIWATAhiroiwata@g.ecc.u-tokyo.ac.jp

2021/4/16

Simple regression analysis

Changesinonevariablemayaffectanother,suchastherelationshipbetweenbreedingandcultivationconditionsandthegrowthofanimalsandplants.Oneofthestatisticalmethodstomodeltherelationshipbetweenthesevariablesisregressionanalysis.Bystatisticallymodelingtherelationshipbetweenvariables,itbecomespossibletounderstandthecausalrelationshipthatexistsbetweenvariables,andtopredictonevariablefromanother.Here,first,wewilldiscusssimpleregressionanalysisthatmodelstherelationshipbetweentwovariablesasa“linearrelationship”.Inthiscase,themechanismofsingleregressionanalysiswillbeexplainedusingtheanalysisofricedata(Zhaoetal.2011,NatureCommunications2:467)asanexample.

First,readthericedatainthesamewayasbefore.Beforeenteringthefollowingcommand,changeyourRworkingdirectorytothedirectory(folder)wherethetwoinputfiles(RiceDiversityPheno.csv,RiceDiversityLine.csv)arelocated.

# this data set was analyzed in Zhao 2011 (Nature Communications 2:467)pheno <- read.csv("RiceDiversityPheno.csv")line <- read.csv("RiceDiversityLine.csv")line.pheno <- merge(line, pheno, by.x = "NSFTV.ID", by.y = "NSFTVID")head(line.pheno)[,1:12]

## NSFTV.ID GSOR.ID IRGC.ID Accession.Name Country.of.origin Latitude## 1 1 301001 To be assigned Agostano Italy 41.871940## 2 3 301003 117636 Ai-Chiao-Hong China 27.902527## 3 4 301004 117601 NSF-TV 4 India 22.903081## 4 5 301005 117641 NSF-TV 5 India 30.472664## 5 6 301006 117603 ARC 7229 India 22.903081## 6 7 301007 To be assigned Arias Indonesia -0.789275## Longitude Sub.population PC1 PC2 PC3 PC4## 1 12.56738 TEJ -0.0486 0.0030 0.0752 -0.0076## 2 116.87256 IND 0.0672 -0.0733 0.0094 -0.0005## 3 87.12158 AUS 0.0544 0.0681 -0.0062 -0.0369## 4 75.34424 AROMATIC -0.0073 0.0224 -0.0121 0.2602## 5 87.12158 AUS 0.0509 0.0655 -0.0058 -0.0378## 6 113.92133 TRJ -0.0293 -0.0027 -0.0677 -0.0085

Prepareanalysisdatabyextractingonlythedatausedforsimpleregressionanalysisfromthereaddata.Hereweanalyzetherelationshipbetweenplantheight(Plant.height)andfloweringtiming(Flowering.time.at.Arkansas).Inaddition,principalcomponentscores(PC1toPC4)

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representingthegeneticbackgroundtobeusedlaterarealsoextracted.Also,removesampleswithmissingvaluesinadvance.

# extract variables for regression analysisdata <- data.frame( height = line.pheno$Plant.height, flower = line.pheno$Flowering.time.at.Arkansas, PC1 = line.pheno$PC1, PC2 = line.pheno$PC2, PC3 = line.pheno$PC3, PC4 = line.pheno$PC4)data <- na.omit(data)head(data)

## height flower PC1 PC2 PC3 PC4## 1 110.9167 75.08333 -0.0486 0.0030 0.0752 -0.0076## 2 143.5000 89.50000 0.0672 -0.0733 0.0094 -0.0005## 3 128.0833 94.50000 0.0544 0.0681 -0.0062 -0.0369## 4 153.7500 87.50000 -0.0073 0.0224 -0.0121 0.2602## 5 148.3333 89.08333 0.0509 0.0655 -0.0058 -0.0378## 6 119.6000 105.00000 -0.0293 -0.0027 -0.0677 -0.0085

First,visualizetherelationshipbetweentwovariables.

# look at the relationship between plant height and flowering timeplot(data$height ~ data$flower)

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Asshownintheabovefigure,theearlierthefloweringtime,theshortertheplantheight,whilethelaterthefloweringtime,thetallertheplantheight.

Let’screateasimpleregressionmodelthatexplainsthevariationinplantheightbythedifferenceinfloweringtiming.

# perform single linear regressionmodel <- lm(height ~ flower, data = data)

Theresultofregressionanalysis(estimatedmodel)isassignedto“model”.Usethefunction“summary”todisplaytheresultofregressionanalysis.

# show the resultsummary(model)

## ## Call:## lm(formula = height ~ flower, data = data)## ## Residuals:## Min 1Q Median 3Q Max ## -43.846 -13.718 0.295 13.409 61.594 ## ## Coefficients:## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 58.05464 6.92496 8.383 1.08e-15 ***## flower 0.67287 0.07797 8.630 < 2e-16 ***## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## ## Residual standard error: 19 on 371 degrees of freedom## Multiple R-squared: 0.1672, Adjusted R-squared: 0.1649 ## F-statistic: 74.48 on 1 and 371 DF, p-value: < 2.2e-16

Iwillexplaintheresultsdisplayedbyexecutingtheabovecommandinorder.

Call:lm(formula=height~flower,data=data)

Thisisarepeatofthecommandyouenteredearlier.Ifyougetthisoutputrightafteryoutypeit,itdoesnotseemtobeusefulinformation.However,ifyoumakemultipleregressionmodelsandcomparethemasdescribedlater,itmaybeusefulbecauseyoucanreconfirmthemodelemployedintheanalysis.Here,assumingthattheplantheightis𝑦! andthefloweringtimingis𝑥! ,theregressionanalysisisperformedwiththemodel

𝑦! = 𝜇 + 𝛽𝑥! + 𝜖!

Asmentionedearlier,𝑥! iscalledindependentvariableorexplanatoryvariable,and𝑦! iscalleddependentvariableorresponsevariable.𝜇and𝛽arecalledparametersoftheregressionmodel,and𝜖! iscallederror.Also,𝜇iscalledpopulationinterceptand𝛽iscalledpopulationregressioncoefficient.

Inaddition,sinceitisnotpossibletodirectlyknowthetruevaluesoftheparameters𝜇and𝛽oftheregressionmodel,estimationisperformedbasedonsamples.Theestimatesoftheparameters𝜇and𝛽,whichareestimatedfromthesample,arecalledsampleinterceptandsampleregressioncoefficient,respectively.Thevaluesof𝜇and𝛽estimatedfromthesamplesaredenotedby𝑚and𝑏,respectively.Since𝑚and𝑏arevaluesestimatedfromthesamples,

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theyarerandomvariablesthatvarydependingonthesamplesselectedbychance.Therefore,itfollowsaprobabilitydistribution.Detailswillbedescribedlater.

Residuals:Min1QMedian3QMax-43.846-13.7180.29513.40961.594

Thisoutputgivesanoverviewofthedistributionofresiduals.Youcanusethisinformationtochecktheregressionmodel.Forexample,themodelassumesthattheexpectedvalue(average)oftheerroris0.Youcancheckwhetherthemedianisclosetoit.Youcanalsocheckwhetherthedistributionissymmetricalaround0,i.e.,whetherthemaximumandminimumorthefirstandthirdquantileshavealmostthesamevalue.Inthisexample,themaximumvalueisslightlylargerthantheminimumvalue,butotherwisenomajorissuesarefound.

Coefficients:EstimateStd.ErrortvaluePr(>|t|)(Intercept)58.054646.924968.3831.08e-15***flower0.672870.077978.630<2e-16***—Signif.codes:0‘’0.001‘’0.01‘’0.05‘.’0.1‘’1

Theestimatesofparameters𝜇and𝛽,i.e.,𝑚and𝑏,andtheirstandarderrors,𝑡valuesand𝑝valuesareshown.Asterisksattheendoofeachlinerepresentsignificancelevels.Onestarrepresents5%,twostars1%,andthreestars0.1%.

Residualstandarderror:19on371degreesoffreedomMultipleR-squared:0.1672,AdjustedR-squared:0.1649F-statistic:74.48on1and371DF,p-value:<2.2e-16

Thefirstlineshowsthestandarddeviationoftheresiduals.Thisisthevaluerepresentedby𝑠,where𝑠"istheestimatedvalueoftheerrorvariance𝜎".Thesecondlineisthedeterminationcoefficient𝑅".Thisindexandtheadjusted𝑅"representhowwelltheregressionexplainthevariationofy.ThethirdlineistheresultoftheFtestthatrepresentsthesignificanceoftheregressionmodel.Itisatestunderthehypothesis(nullhypothesis)thatallregressioncoefficientsare0,andifthis𝑝valueisverysmall,thenullhypothesisisrejectedandthealternativehypothesis(regressioncoefficientisnot0)istakentobeadopted.

Let’slookattheresultsofregressionanalysisgraphically.First,drawascatterplotanddrawaregressionline.

# again, plot the two variablesplot(data$height ~ data$flower)abline(model, col = "red")

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Next,calculateandplotthevalueof𝑦whenthedataisfittedtotheregressionmodel.

# calculate fitted valuesheight.fit <- fitted(model)plot(data$height ~ data$flower)abline(model, col = "red")points(data$flower, height.fit, pch = 3, col = "green")

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Thevaluesof𝑦calculatedbyfittingthemodelalllieonastraightline.

Anobservedvalue𝑦isexpressedasthesumofthevariationexplainedbytheregressionmodelandtheerrorwhichisnotexplainedbytheregression.Let’svisualizetheerrorinthefigureandchecktherelationship.

# plot residualsplot(data$height ~ data$flower)abline(model, col = "red")points(data$flower, height.fit, pch = 3, col = "green")segments(data$flower, height.fit, data$flower, height.fit + resid(model), col = "gray")

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Thevalueof𝑦isexpressedasthesumofthevaluesof𝑦calculatedbyfittingthemodel(greenpoints)andtheresidualsofthemodel(graylinesegments)

Let’susearegressionmodeltopredict𝑦for𝑥 = (60,80, . . . ,140),whicharenotactuallyobserved.

# predict unknown dataheight.pred <- predict(model, data.frame(flower = seq(60, 140, 20)))plot(data$height ~ data$flower)abline(model, col = "red")points(data$flower, height.fit, pch = 3, col = "green")segments(data$flower, height.fit, data$flower, height.fit + resid(model), col = "gray")points(seq(60, 140, 20), height.pred, pch = 2, col = "blue")

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Allthepredictedvalueswilllocateagainontheline.

Quiz 1

Now,let’ssolveapracticequestionhere.Practicequestionswillbepresentedinthelecture.

Gotohttps://www.menti.com/ie9s1dzmdzandtypeinthenumberIwilltellyouinthelecture.Then,registeryournicknameandwaitforthequiztostart.

Method for calculating the parameters of a regression model

Herewewillexplainhowtocalculatearegressionmodel.Also,let’scalculatetheregressioncoefficientswhileactuallyusingtheRcommand.

Asmentionedearlier,thesimpleregressionmodelis

𝑦! = 𝑚 + 𝑏𝑥! + 𝜖!

Therearevariouscriteriaforwhatisconsidered“optimal”,buthereweconsiderminimizingtheerror𝜖! acrossthedata.Sinceerrorscantakebothpositiveandnegativevalues,errorscancelseachotherintheirsimplesum.So,weconsiderminimizingthesumofsquarederror(SSE).Thatis,consider𝑚and𝑏thatminimizethefollowingequation:

𝑆𝑆𝐸 =:𝜖!"#

!$%

=:(#

!$%

𝑦! − (𝑚 + 𝑏𝑥!))"

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Equation(1)

ThefollowingfigureshowsthechangeinSSEforvariousvaluesof𝑚and𝑏.Thecommandstodrawthefigureisalittlecomplicated,buttheyareasfollows:

#visualize the plane for optimizationx <- data$flowery <- data$heightm <- seq(0, 100, 1)b <- seq(0, 2, 0.02)sse <- matrix(NA, length(m), length(b))for(i in 1:length(m)) { for(j in 1:length(b)) { sse[i, j] <- sum((y - m[i] - b[j] * x)^2) }}persp(m, b, sse, col = "green")

Drawthegraphusing“plotly”package

# draw the figure with plotlydf <- data.frame(m, b, sse) plot_ly(data = df, x = ~m, y = ~b, z = ~sse) %>% add_surface()

Itshouldbenotedthatatthepointwhere𝑆𝑆𝐸becomestheminimuminFigure3,𝑆𝑆𝐸shouldnotchange(theslopeofthetangentiszero)evenwhen𝜇or𝛽changesslightly.Therefore,thecoordinatesoftheminimumpointcanbedeterminedbypartiallydifferentiatingtheequation(1)with𝜇and𝛽,andsettingthevaluetozero.Thatis,

𝜕𝑆𝑆𝐸𝜕𝑚 = 0,

𝜕𝑆𝑆𝐸𝜕𝑏 = 0

Weshouldobtainthevaluesof𝜇and𝛽tosatisfythese.Themethodofcalculatingtheparametersofaregressionmodelthroughminimizingthesumofsquaresoferrorsinthiswayiscalledtheleastsquaresmethod.

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Notethat𝜇minimizingSSEis

𝜕𝑆𝑆𝐸𝜕𝑚 = −2:(

#

!$%

𝑦! −𝑚 − 𝑏𝑥!) = 0

⇔:𝑦!

#

!$%

− 𝑛𝑚 − 𝑏:𝑥!

#

!$%

= 0

⇔𝑚 =∑ 𝑦!#!$%

𝑛 − 𝑏∑ 𝑥!#!$%

𝑛 = 𝑦 − 𝑏𝑥

Also,𝛽minimizing𝑆𝑆𝐸is

𝜕𝑆𝑆𝐸𝜕𝑏 = −2:𝑥!

#

!$%

(𝑦! −𝑚 − 𝑏𝑥!) = 0

⇔:𝑥!

#

!$%

𝑦! −𝑚:𝑥!

#

!$%

− 𝑏:𝑥!"#

!$%

= 0

⇔:𝑥!

#

!$%

𝑦! − 𝑛(𝑦 − 𝑏𝑥)𝑥 − 𝑏:𝑥!"#

!$%

= 0

⇔:𝑥!

#

!$%

𝑦! − 𝑛𝑥𝑦 − 𝑏(:𝑥!"#

!$%

− 𝑛𝑥") = 0

⇔ 𝑏 =∑ 𝑥!#!$% 𝑦! − 𝑛𝑥𝑦∑ 𝑥!"#!$% − 𝑛𝑥"

=𝑆𝑆𝑋𝑌𝑆𝑆𝑋

Here,𝑆𝑆𝑋𝑌and𝑆𝑆𝑋aresumofproductsofdeviationin𝑥and𝑦anddeviationthesumofsquaresofdeviationin𝑥,respectively.

𝑆𝑆𝑋𝑌 =:(#

!$%

𝑥! − 𝑥)(𝑦! − 𝑦)

=:𝑥!

#

!$%

𝑦! − 𝑥:𝑦!

#

!$%

− 𝑦:𝑥!

#

!$%

+ 𝑛𝑥𝑦

=:𝑥!

#

!$%

𝑦! − 𝑛𝑥𝑦 − 𝑛𝑦𝑥 + 𝑛𝑥𝑦

=:𝑥!

#

!$%

𝑦! − 𝑛𝑥𝑦

𝑆𝑆𝑋 =:(#

!$%

𝑥! − 𝑥)"

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=:𝑥!"#

!$%

− 2𝑥:𝑥!

#

!$%

− 𝑛𝑥"

=:𝑥!"#

!$%

− 2𝑛𝑥" + 𝑛𝑥"

=:𝑥!"#

!$%

− 𝑛𝑥"

Thevaluesof𝑚and𝑏minimizing𝑆𝑆𝐸aretheestimatesoftheparameters,andlettheestimatesberepresentedsimplyby𝑚and𝑏.Thatis,

𝑏 =𝑆𝑆𝑋𝑌𝑆𝑆𝑋

𝑚 = 𝑦 − 𝑏𝑥

Nowlet’scalculatetheregressioncoefficientsbasedontheaboveequation.First,calculatethesumofproductsfdeviationandthesumofsquaresofdeviation.

# calculate sum of squares (ss) of x and ss of xyn <- length(x)ssx <- sum(x^2) - n * mean(x)^2ssxy <- sum(x * y) - n * mean(x) * mean(y)

Firstwecalculatetheslope𝑏.

# calculate bb <- ssxy / ssxb

## [1] 0.6728746

Thencalculatetheintercept𝑚

# calculate mm <- mean(y) - b * mean(x)m

## [1] 58.05464

Let’sdrawaregressionlinebasedonthecalculatedestimates.

# draw scatter plot and regression line plot(y ~ x)abline(m, b)

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Let’smakesurethatthesameregressionlinewasobtainedasthefunctionlmwhichweusedearlier.

Notethatoncetheregressionparameters𝜇and𝛽areestimated,itispossibletocalculate𝑦C! ,whichisthevalueof𝑦correspondingtoagiven𝑥! .Thatis,

𝑦&D = 𝑚 + 𝑏𝑥!

Thismakesitpossibletocalculatethevalueof𝑦whenthemodelisfittedtotheobserved𝑥,ortopredict𝑦ifonlythevalueof𝑥isknown.Here,let’scalculatethevalueof𝑦whenthemodelisfittedtotheobserved𝑥,anddrawscatterpointsonthefiguredrawnearlier.

# calculate fitted valuesy.hat <- m + b * xlim <- range(c(y, y.hat))plot(y, y.hat, xlab = "Observed", ylab = "Fitted", xlim = lim, ylim = lim)abline(0, 1)

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Let’scalculatethecorrelationcoefficientbetweenthetwotofindthedegreeofagreementbetweentheobservedandfittedvalues.

# calculate correlation between observed and fitted valuescor(y, y.hat)

## [1] 0.408888

Infact,thesquareofthiscorrelationcoefficientistheproportionofthevariationof𝑦explainedbytheregression(coefficientofdetermination,𝑅"value).Let’scomparethesetwostatistics.

# compare the square of the correlation and R2 cor(y, y.hat)^2

## [1] 0.1671894

summary(model)

## ## Call:## lm(formula = height ~ flower, data = data)## ## Residuals:## Min 1Q Median 3Q Max ## -43.846 -13.718 0.295 13.409 61.594 ##

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## Coefficients:## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 58.05464 6.92496 8.383 1.08e-15 ***## flower 0.67287 0.07797 8.630 < 2e-16 ***## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## ## Residual standard error: 19 on 371 degrees of freedom## Multiple R-squared: 0.1672, Adjusted R-squared: 0.1649 ## F-statistic: 74.48 on 1 and 371 DF, p-value: < 2.2e-16

Quiz 2

Now,let’ssolveapracticequestionhere.Practicequestionswillbepresentedinthelecture.

Ifyouclosethepageofthequiz,gotohttps://www.menti.com/ie9s1dzmdzandtypeinthenumberIwilltellyouinthelecture.

Significance test of a regression model

Whenthelinearrelationshipbetweenvariablesisstrong,aregressionlinefitwelltotheobservations,andtherelationshipbetweenbothvariablescanbewellmodeledbyaregressionline.However,whenalinearrelationshipbetweenvariablesisnotclear,modelingwitharegressionlinedoesnotworkwell.Here,asamethodtoobjectivelyconfirmthegoodnessoffitoftheestimatedregressionmodel,wewillexplainatestusinganalysisofvariance.

First,let’sgobacktothesimpleregressionagain.

model <- lm(height ~ flower, data = data)

Thesignificanceoftheobtainedregressionmodelcanbetestedusingthefunctionanova.

# analysis of variance of regressionanova(model)

## Analysis of Variance Table## ## Response: height## Df Sum Sq Mean Sq F value Pr(>F) ## flower 1 26881 26881.5 74.479 < 2.2e-16 ***## Residuals 371 133903 360.9 ## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Asaresultoftheanalysisofvariance,thetermoffloweringtimeishighlysignificant(𝑝 <0.001),andthegoodnessoffitoftheregressionmodelthatthefloweringtiminginfluencesplantheightisconfirmed.

Analysisofvarianceforregressionmodelsinvolvesthefollowingcalculations:Firstofall,“Sumofsquaresexplainedbyregression”canbecalculatedasfollows.

𝑆𝑆𝑅 =:(#

!$%

𝑦C! − 𝑦)"

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=:(#

!$%

𝑚+ 𝑏𝑥! − (𝑚 + 𝑏𝑥))"

= 𝑏":(#

!$%

𝑥! − 𝑥)"

= 𝑏" ⋅ 𝑆𝑆𝑋 = 𝑏 ⋅ 𝑆𝑆𝑋𝑌

Also,thesumofsquaresofdeviationfromthemeanoftheobservedvalues𝑦isexpressedasthesumofthesumofsquares𝑆𝑆𝑅explainedbytheregressionandtheresidualsumofsquares𝑆𝑆𝐸.Thatis,

𝑆𝑆𝑌 =:(#

!$%

𝑦! − 𝑦)"

=:(#

!$%

𝑦! − 𝑦C! + 𝑦C! − 𝑦)"

=:(#

!$%

𝑦! − 𝑦C!)" +:(#

!$%

𝑦C! − 𝑦)"

= 𝑆𝑆𝐸 + 𝑆𝑆𝑅

∵ 2:(#

!$%

𝑦! − 𝑦C!)(𝑦C! − 𝑦)

= 2:(#

!$%

𝑦! −𝑚 − 𝑏𝑥!)(𝑚 + 𝑏𝑥! − (𝑚 + 𝑏𝑥))

= 2𝑏:(#

!$%

𝑦! − (𝑦 − 𝑏𝑥) − 𝑏𝑥!)(𝑥! − 𝑥)

= 2𝑏:(#

!$%

𝑦! − 𝑦 − 𝑏(𝑥! − 𝑥))(𝑥! − 𝑥)

= 2𝑏(𝑆𝑆𝑋𝑌 − 𝑏 ⋅ 𝑆𝑆𝑋) = 0

Let’sactuallycalculateitusingtheaboveequation.First,calculate𝑆𝑆𝑅and𝑆𝑆𝐸.

# calculate sum of squares of regression and errorssr <- b * ssxyssr

## [1] 26881.49

ssy <- sum(y^2) - n * mean(y)^2sse <- ssy - ssrsse

## [1] 133903.2

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Next,calculatethemeansquares,whichisofthesumofsquaresdividedbythedegreesoffreedom.

# calculate mean squares of regression and errormsr <- ssr / 1msr

## [1] 26881.49

mse <- sse / (n - 2)mse

## [1] 360.9251

Finally,themeansquareoftheregressionisdividedbythemeansquareoftheerrortocalculatethe𝐹value.Furthermore,calculatethe𝑝valuecorrespondingtothecalculatedFvalue.

# calculate F valuef.value <- msr / msef.value

## [1] 74.47943

# calculate p value for the F value1 - pf(f.value, 1, n - 2)

## [1] 2.220446e-16

Theresultsobtainedareinagreementwiththeresultscalculatedearlierusingthefunctionanova.

Theresultsofregressionanalysisareincludedintheresultsofregressionanalysisdisplayedusingthefunction“summary”.

# check the summary of the result of regression analysis summary(model)

## ## Call:## lm(formula = height ~ flower, data = data)## ## Residuals:## Min 1Q Median 3Q Max ## -43.846 -13.718 0.295 13.409 61.594 ## ## Coefficients:## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 58.05464 6.92496 8.383 1.08e-15 ***## flower 0.67287 0.07797 8.630 < 2e-16 ***## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## ## Residual standard error: 19 on 371 degrees of freedom## Multiple R-squared: 0.1672, Adjusted R-squared: 0.1649 ## F-statistic: 74.48 on 1 and 371 DF, p-value: < 2.2e-16

“Residualstandarderror”isthesquarerootofthemeansquareoftheresidual.

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# square root of msesqrt(mse)

## [1] 18.99803

“MultipleR-squared”(𝑅")isavaluecalledthecoefficientofdetermination,whichistheratioof𝑆𝑆𝑅to𝑆𝑆𝑌.

# R squared ssr / ssy

## [1] 0.1671894

“AdjustedR-squared”(𝑅'()" )isavaluecalledtheadjustedcoefficientofdetermination,whichcanbecalculatedasfollows.

# adjusted R squared(ssy / (n - 1) - mse) / (ssy / (n - 1))

## [1] 0.1649446

Also,“F-statistic”matchesthe𝐹valueandits𝑝valuewhichareexpressedastheeffectoffloweringtimeintheanalysisofvariance.Inaddition,thetvaluecalculatedfortheregressioncoefficientofthefloweringtimetermissquaredtoobtainthe𝐹value(8.6302=74.477).

𝑅"and𝑅'()" canalsobeexpressedusing𝑆𝑆𝑅,𝑆𝑆𝑌,and𝑆𝑆𝐸asfollows.

𝑅" =𝑆𝑆𝑅𝑆𝑆𝑌 = 1 −

𝑆𝑆𝐸𝑆𝑆𝑌

𝑅'()" = 1 −𝑛 − 1𝑛 − 𝑝 ⋅

𝑆𝑆𝐸𝑆𝑆𝑌

Here,𝑝isthenumberofparametersincludedinthemodel,and𝑝 = 2forasimpleregressionmodel.Itcanbeseenthat𝑅'()" hasalargeramountofadjustment(theresidualsumofsquaresisunderestimated)asthenumberofparametersincludedinthemodelincreases.

Quiz 3

Now,let’ssolveapracticequestionhere.Practicequestionswillbepresentedinthelecture.

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Distribution that estimated value of regression coefficient follows

Asmentionedearlier,theestimates𝑏and𝑚oftheregressioncoefficients𝜇and𝛽arevaluesestimatedfromsamplesandarerandomvariablesthatdependonthesampleschosenbychance.Thus,estimates𝑏and𝑚haveprobabilisticdistributions.Hereweconsiderthedistributionsthattheestimatesfollow.

Theestimate𝑏followsthenormaldistribution:

𝑏 ∼ 𝑁(𝛽,𝜎"

𝑆𝑆𝑋)

Here,𝜎"istheresidualvariance𝜎" = 𝑉𝑎𝑟(𝑦!) = 𝑉𝑎𝑟(𝑒!).

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Theesimate𝑚followsthenormaldistribution:

𝑚 ∼ 𝑁(𝜇, 𝜎"[1𝑛 +

𝑥"

𝑆𝑆𝑋])

Althoughthetruevalueoftheerrorvariance𝜎"isunknown,itcanbereplacedbytheresidualvariance𝑠".Thatis,

𝑠" =𝑆𝑆𝐸𝑛 − 2

Thisvalueisthemeansquareoftheresidualscalculatedduringtheanalysisofvariance.

Atthistime,statisticson𝑏

𝑡 =𝑏 − 𝑏*𝑠/√𝑆𝑆𝑋

followsthetdistributionwithn–2degreesoffreedomunderthenullhypothesis:

𝐻*: 𝛽 = 𝑏*

First,testfornullhypothesis𝐻*: 𝛽 = 0for𝑏.

# test beta = 0t.value <- (b - 0) / sqrt(mse/ssx)t.value

## [1] 8.630147

Thisstatisticfollowsthetdistributionwith371degreesoffreedomunderthenullhypothesis.Drawagraphofthedistribution

# visualize the t distribution under H0s <- seq(-10, 10, 0.2)plot(s, dt(s, n - 2), type = "l")abline(v = t.value, col = "green")abline(v = - t.value, col = "gray")

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Fromthedistributionthatfollowsunderthenullhypothesis,thevalueof𝑡thatisbeingobtainedappearstobelarge.Let’sperforma𝑡test.Notethatitisatwo-tailedtest.

# perform t test2 * (1 - pt(abs(t.value), n - 2)) # two-sided test

## [1] 2.220446e-16

Theresultsofthistestwerealreadydisplayedasregressionanalysisresults.

# check the summary of the modelsummary(model)

## ## Call:## lm(formula = height ~ flower, data = data)## ## Residuals:## Min 1Q Median 3Q Max ## -43.846 -13.718 0.295 13.409 61.594 ## ## Coefficients:## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 58.05464 6.92496 8.383 1.08e-15 ***## flower 0.67287 0.07797 8.630 < 2e-16 ***## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

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## ## Residual standard error: 19 on 371 degrees of freedom## Multiple R-squared: 0.1672, Adjusted R-squared: 0.1649 ## F-statistic: 74.48 on 1 and 371 DF, p-value: < 2.2e-16

Thehypothesistestperformedabovecanbeperformedforany𝛽*.Forexample,let’stestfornullhypothesis𝐻*: 𝛽 = 0.5.

# test beta = 0.5t.value <- (b - 0.5) / sqrt(mse/ssx)t.value

## [1] 2.217253

2 * (1 - pt(abs(t.value), n - 2))

## [1] 0.02721132

Theresultissignificantatthe5%level.

Nowletustestfor𝑚.First,let’stestthenullhypothesis𝐻*: 𝜇 = 0.

# test mu = 0t.value <- (m - 0) / sqrt(mse * (1/n + mean(x)^2 / ssx))t.value

## [1] 8.383389

2 * (1 - pt(abs(t.value), n - 2))

## [1] 1.110223e-15

Thisresultwasalsoalreadycalculated.

# check the summary of the model againsummary(model)

## ## Call:## lm(formula = height ~ flower, data = data)## ## Residuals:## Min 1Q Median 3Q Max ## -43.846 -13.718 0.295 13.409 61.594 ## ## Coefficients:## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 58.05464 6.92496 8.383 1.08e-15 ***## flower 0.67287 0.07797 8.630 < 2e-16 ***## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## ## Residual standard error: 19 on 371 degrees of freedom## Multiple R-squared: 0.1672, Adjusted R-squared: 0.1649 ## F-statistic: 74.48 on 1 and 371 DF, p-value: < 2.2e-16

(Itmaybeduetotheroundingerrorthatthepvaluedoesnotmatchcompletely)

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Finally,let’stestforthenullhypothesis𝐻*: 𝜇 = 70.

# test mu = 70t.value <- (m - 70) / sqrt(mse * (1/n + mean(x)^2 / ssx))t.value

## [1] -1.724971

2 * (1 - pt(abs(t.value), n - 2))

## [1] 0.08536545

Theresultwasnotsignificantevenatthe5%level.

Quiz 4

Now,let’ssolveapracticequestionhere.Practicequestionswillbepresentedinthelecture.

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Confidence intervals for regression coefficients and fitted values

Function“predict”hasvariousfunctions.First,let’susethefunctionwiththeestimatedregressionmodel.Then,thevaluesofywhenthemodelfittedtoobserveddataarecalculated.Thevalues𝑦Careexactlythesameascalculatedbythefunction“fitted”.

# fitted valuespred <- predict(model)head(pred)

## 1 2 3 4 5 6 ## 108.5763 118.2769 121.6413 116.9312 117.9966 128.7065

head(fitted(model))

## 1 2 3 4 5 6 ## 108.5763 118.2769 121.6413 116.9312 117.9966 128.7065

Bysettingtheoptions“interval”and“level”,youcancalculatetheconfidenceinterval(the95confidenceintervalatthedefaultsetting)ofyatthespecifiedsignificancelevelwhenfittingthemodel.

# calculate confidence intervalpred <- predict(model, interval = "confidence", level = 0.95)head(pred)

## fit lwr upr## 1 108.5763 105.8171 111.3355## 2 118.2769 116.3275 120.2264## 3 121.6413 119.4596 123.8230## 4 116.9312 114.9958 118.8665## 5 117.9966 116.0540 119.9391## 6 128.7065 125.4506 131.9623

Let’svidualizethe95%confidenceintervalofyusingthefunction“predict”.

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# draw confidence bandspred <- data.frame(flower = 50:160)pc <- predict(model, interval = "confidence", newdata = pred)plot(data$height ~ data$flower)matlines(pred$flower, pc, lty = c(1, 2, 2), col = "red")

Next,wewillconsiderthecaseofpredictingunknowndata.Letususethepredictfunctiontodrawthe95%confidenceintervalofthepredictedvalue,i.e.,thevalueof𝑦thatwouldbeobservedfortheunknowndata,i.e.,thepredictedvalue𝑦X,issubjecttoadditionalvariationduetoerror.

pc <- predict(model, interval= "prediction", newdata = pred)plot(data$height ~ data$flower)matlines(pred$flower, pc, lty = c(1, 2, 2), col = "green")

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Withtheaddederror,thevariabilityofthepredicted𝑦Xisgreaterthantheestimated𝑦C.

Notethatforaparticular𝑥,theconfidenceintervalsfortheestimatedy,i.e.,𝑦C,andthepredictedy,i.e.,𝑦X,canbefoundasfollows.Here,wefindthe99%confidenceintervalwhen𝑥 =120.

# estimate the confidence intervals for the estimate and prediction of y pred <- data.frame(flower = 120)predict(model, interval = "confidence", newdata = pred, level = 0.99)

## fit lwr upr## 1 138.7996 131.8403 145.7589

predict(model, interval = "prediction", newdata = pred, level = 0.99)

## fit lwr upr## 1 138.7996 89.12106 188.4781

Quiz 5

Now,let’ssolveapracticequestionhere.Practicequestionswillbepresentedinthelecture.

Ifyouclosethepageofthequiz,gotohttps://www.menti.com/ie9s1dzmdzandtypeinthenumberIwilltellyouinthelecture.

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Polynomial regression model and multiple regression model

Sofar,wehaveappliedtothedataaregressionmodelthatrepresentstherelationshipbetweenthetwovariableswithastraightline.Let’sextendtheregressionmodelabit.

First,let’sperformregressionbyamethodcalledpolynomialregression.Inpolynomialregression,

𝑦! = 𝜇 + 𝛽%𝑥! + 𝛽"𝑥!" +⋯+ 𝛽+𝑥!+ + 𝜖!

Inthisway,regressionisperformedusingthesecondorhigherordertermsof𝑥.First,let’sperformregressionusingthefirstandsecondtermsof𝑥.

# polynomial regressionmodel.quad <- lm(height ~ flower + I(flower^2), data = data)summary(model.quad)

## ## Call:## lm(formula = height ~ flower + I(flower^2), data = data)## ## Residuals:## Min 1Q Median 3Q Max ## -39.57 -13.60 0.97 12.91 64.83 ## ## Coefficients:## Estimate Std. Error t value Pr(>|t|) ## (Intercept) -29.082326 27.019440 -1.076 0.282473 ## flower 2.662663 0.601878 4.424 1.28e-05 ***## I(flower^2) -0.011130 0.003339 -3.333 0.000945 ***## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## ## Residual standard error: 18.74 on 370 degrees of freedom## Multiple R-squared: 0.1915, Adjusted R-squared: 0.1871 ## F-statistic: 43.81 on 2 and 370 DF, p-value: < 2.2e-16

Itcanbeseenthattheproportionofvariationof𝑦(coefficientofdetermination𝑅")explainedbythepolynomialregressionmodelislargerthanthatofthesimpleregressionmodel.

Althoughthiswillbementionedlater,youshouldnotjudgethatthepolynomialregressionmodelisexcellentonlywiththisvalue.Thisisbecauseapolynomialregressionmodelhasmoreparametersthanasimpleregressionmodel,andyouhavemoreflexibilitywhenfittingthemodeltodata.Itiseasytoimprovethefitofthemodeltothedatabyincreasingtheflexibility.Inextremecases,themodelcanbecompletelyfittedtothedatawithasmanyparametersasthesizeofdata(Inthatcase,thecoefficientpfdetermination𝑅"completelymatches1).Therefore,carefulselectionofsomestatisticalcriteriaisrequiredwhenselectingthebestmodel.Thiswillbediscussedlater.

# plot(data$height ~ data$flower)pred <- data.frame(flower = 50:160)pc <- predict(model.quad, interval = "confidence", newdata = pred)plot(data$height ~ data$flower)matlines(pred$flower, pc, lty = c(1, 2, 2), col = "red")

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Whenthetimingoffloweringisover120daysaftersowing,itcanbeseenthatthereliabilityofthemodelislow.

Nowlet’sdrawtheresultofpolynomialregressionwithconfidenceintervals.

# compare predicted and observed valueslim <- range(c(data$height, fitted(model), fitted(model.quad)))plot(data$height, fitted(model), xlab = "Observed", ylab = "Expected", xlim = lim, ylim = lim)points(data$height, fitted(model.quad), col = "red")abline(0, 1)

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Theabovefigurerepresentsrelationshipbetweenfittedvalueandobservedvalueofthesimpleregressionmodel(black)andthesecond-orderpolynomialmodel(red).

Let’stestiftheimprovementintheexplanatorypowerofthesecond-orderpolynomialmodelisstatisticallysignificant.Thesignificanceistestedwith𝐹testwhetherthedifferencebetweentheresidualsumofsquaresofthetwomodelsissufficientlylargecomparedtotheresidualsumofsquaresofthemodelcontainingtheother(here,Model2containsModel1).

# compare error variance between two modelsanova(model, model.quad)

## Analysis of Variance Table## ## Model 1: height ~ flower## Model 2: height ~ flower + I(flower^2)## Res.Df RSS Df Sum of Sq F Pr(>F) ## 1 371 133903 ## 2 370 129999 1 3903.8 11.111 0.0009449 ***## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Theresultsshowthatthedifferenceinresidualvariancebetweenthetwomodelsishighlysignificant(𝑝 < 0.001).Inotherwords,Model2hassignificantlymoreexplanatorypowerthanModel1.

Nowlet’sfitathird-orderpolynomialregressionmodelandtestifitissignificantlymoredescriptivethanasecond-ordermodel.

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# extend polynormial regression model to a higher dimensional one...model.cube <- lm(height ~ flower + I(flower^2) + I(flower^3), data = data)summary(model.cube)

## ## Call:## lm(formula = height ~ flower + I(flower^2) + I(flower^3), data = data)## ## Residuals:## Min 1Q Median 3Q Max ## -39.699 -13.705 1.031 13.240 65.840 ## ## Coefficients:## Estimate Std. Error t value Pr(>|t|) ## (Intercept) -1.001e+02 8.541e+01 -1.172 0.2419 ## flower 5.029e+00 2.765e+00 1.818 0.0698 .## I(flower^2) -3.664e-02 2.929e-02 -1.251 0.2118 ## I(flower^3) 8.898e-05 1.015e-04 0.877 0.3813 ## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## ## Residual standard error: 18.75 on 369 degrees of freedom## Multiple R-squared: 0.1931, Adjusted R-squared: 0.1866 ## F-statistic: 29.44 on 3 and 369 DF, p-value: < 2.2e-16

# compare error variance between two modelsanova(model.quad, model.cube)

## Analysis of Variance Table## ## Model 1: height ~ flower + I(flower^2)## Model 2: height ~ flower + I(flower^2) + I(flower^3)## Res.Df RSS Df Sum of Sq F Pr(>F)## 1 370 129999 ## 2 369 129729 1 270.17 0.7685 0.3813

The3rd-ordermodelhasaslightlybetterexplanatorypowerthanthe2nd-ordermodel.However,thedifferenceisnotstatisticallysignificant.Inotherwords,itturnsoutthatextendingasecond-ordermodeltoathird-ordermodelisnotagoodidea.

Finally,let’sapplythemultiplelinearregressionmodel:

𝑦! = 𝜇 + 𝛽%𝑥%! + 𝛽"𝑥"! +⋯+ 𝛽+𝑥+! + 𝜖!

Inthisway,regressionisperformedusingmultipleexplanatoryvariables(𝑥%! , 𝑥"! , . . . , 𝑥+!).Inthefirstlecture,Iconfirmedinthegraphthattheheightvariesdependingonthedifferenceingeneticbackground.Here,wewillcreateamultipleregressionmodelthatexplainsplantheightusinggeneticbackgrounds(PC1toPC4)expressedasthescoresoffourprincipalcomponents.

# multi-linear regression with genetic background model.wgb <- lm(height ~ PC1 + PC2 + PC3 + PC4, data = data)summary(model.wgb)

## ## Call:## lm(formula = height ~ PC1 + PC2 + PC3 + PC4, data = data)

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## ## Residuals:## Min 1Q Median 3Q Max ## -45.89 -11.65 0.15 11.05 72.12 ## ## Coefficients:## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 117.2608 0.8811 133.080 < 2e-16 ***## PC1 181.6572 18.2977 9.928 < 2e-16 ***## PC2 83.5334 17.9920 4.643 4.79e-06 ***## PC3 -88.6432 18.1473 -4.885 1.55e-06 ***## PC4 122.1351 18.2476 6.693 8.16e-11 ***## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## ## Residual standard error: 17 on 368 degrees of freedom## Multiple R-squared: 0.3388, Adjusted R-squared: 0.3316 ## F-statistic: 47.14 on 4 and 368 DF, p-value: < 2.2e-16

anova(model.wgb)

## Analysis of Variance Table## ## Response: height## Df Sum Sq Mean Sq F value Pr(>F) ## PC1 1 28881 28881.3 99.971 < 2.2e-16 ***## PC2 1 5924 5924.2 20.506 8.040e-06 ***## PC3 1 6723 6723.2 23.272 2.063e-06 ***## PC4 1 12942 12942.3 44.799 8.163e-11 ***## Residuals 368 106314 288.9 ## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Youcanseethatthecoefficientofdeterminationoftheregressionmodelishigherthanthatofthepolynomialregressionmodel.Theresultsofanalysisofvarianceshowthatallprincipalcomponentsaresignificantandneedtobeincludedintheregression.

Finally,let’scombinethepolynomialregressionmodelwiththemultipleregressionmodel.

# multi-linear regression with all informationmodel.all <- lm(height ~ flower + I(flower^2) + PC1 + PC2 + PC3 + PC4, data = data)summary(model.all)

## ## Call:## lm(formula = height ~ flower + I(flower^2) + PC1 + PC2 + PC3 + ## PC4, data = data)## ## Residuals:## Min 1Q Median 3Q Max ## -37.589 -10.431 0.542 10.326 65.390 ## ## Coefficients:## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 25.739160 24.725955 1.041 0.29857 ## flower 1.633185 0.543172 3.007 0.00282 **

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## I(flower^2) -0.006598 0.002974 -2.219 0.02713 * ## PC1 141.214491 18.547296 7.614 2.29e-13 ***## PC2 83.552448 17.231568 4.849 1.84e-06 ***## PC3 -45.310663 18.647979 -2.430 0.01559 * ## PC4 119.638954 17.369423 6.888 2.48e-11 ***## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1## ## Residual standard error: 16.17 on 366 degrees of freedom## Multiple R-squared: 0.4045, Adjusted R-squared: 0.3947 ## F-statistic: 41.43 on 6 and 366 DF, p-value: < 2.2e-16

# compare error varianceanova(model.all, model.wgb)

## Analysis of Variance Table## ## Model 1: height ~ flower + I(flower^2) + PC1 + PC2 + PC3 + PC4## Model 2: height ~ PC1 + PC2 + PC3 + PC4## Res.Df RSS Df Sum of Sq F Pr(>F) ## 1 366 95753 ## 2 368 106314 -2 -10561 20.184 4.84e-09 ***## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Theeffectofthegeneticbackgroundonplantheightisverylarge,butitcanalsobeseenthatthemodel’sexplanatorypowerimprovesiftheeffectoffloweringtimingisalsoadded.

Lastly,let’scomparethefirstregressionmodelandthelastcreatedmultipleregressionmodelbyplottingthescatteroftheobservedvalueandthefittedvalue.

# compare between the simplest and final modelslim <- range(data$height, fitted(model), fitted(model.all))plot(data$height, fitted(model), xlab = "Observed", ylab = "Fitted", xlim = lim, ylim = lim)points(data$height, fitted(model.all), col = "red")abline(0,1)

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Asaresult,wecanseethattheexplanatorypowerofthemodelissignificantlyimprovedbyconsideringthegeneticbackgroundandthesecondorderterms.However,ontheotherhand,itcanalsobeseenthatthetwovarietiesandlineswhosefloweringtimingislate(after180days)cannotbesufficientlyexplainedevenbythefinallyobtainedmodel.Theremayberoomtoimprovethemodel,suchasaddingnewfactorsasindependentvariables.

Quiz 6

Now,let’ssolveapracticequestionhere.Practicequestionswillbepresentedinthelecture.

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Experimental design and analysis of variance

Whentryingtodrawconclusionsbasedonexperimentalresults,itisalwaysthepresenceoferrorsintheobservedvalues.Errorsareinevitablenomatterhowprecisetheexperimentis,especiallyinfieldexperiments,errorsarecausedbysmallenvironmentalvariationsinthefield.Therefore,experimentaldesignisamethoddevisedtoobtainobjectiveconclusionswithoutbeingaffectedbyerrors.

Firstofall,whatismostimportantinplanningexperimentsistheFisher’sthreeprinciples:

1. Replication:Inordertobeabletoperformstatisticaltestsonexperimentalresults,werepeatthesameprocess.Forexample,evaluateonevarietymultipletimes.Theexperimentalunitequivalenttoonereplicationiscalledaplot.

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2. Randomization:Anoperationthatmakestheeffectoferrorsrandomiscalledrandomization.Forexample,inthefieldtestexample,varietiesarerandomlyassignedtoplotsinthefieldusingdiceorrandomnumbers.

3. Localcontrol:Localcontrolmeansdividingthefieldintoblocksandmanagingtheenvironmentalconditionsineachblocktobeashomogeneousaspossible.Intheexampleofthefieldtest,thegroupedareaofthefieldisdividedintosmallunitscalledblockstomakethecultivationenvironmentintheblockashomogeneousaspossible.Itiseasiertohomogenizeeachblockratherthanhomogenizingthecultivationenvironmentofthewholefield.

Theexperimentalmethodofdividingthefieldintoseveralblocksandmakingthecultivationenvironmentashomogeneousaspossibleintheblocksiscalledtherandomizedblockdesign.Intherandomizedblockdesign,thefieldisdividedintoblocks,andvarietiesarerandomlyassignedwithineachblock.Thenumberofblocksisequaltothenumberofreplications.

Next,Iwillexplainthemethodofstatisticaltestintherandomizedblockdesignthroughasimplesimulation.First,let’ssetthe“seed”oftherandomnumberbeforestartingthesimulation.Arandomseedisasourcevalueforgeneratingpseudorandomnumbers.

# set a seed for random number generationset.seed(12)

Let’sstartthesimulation.Here,considerafieldwhere16plotsarearrangedin4×4.Andthinkaboutthesituationthatthereisaslopeofthesoilfertilityinthefield.

# The blocks have unequal fertility among themfield.cond <- matrix(rep(c(4,2,-2,-4), each = 4), nrow = 4)field.cond

## [,1] [,2] [,3] [,4]## [1,] 4 2 -2 -4## [2,] 4 2 -2 -4## [3,] 4 2 -2 -4## [4,] 4 2 -2 -4

However,itisassumedthatthereisaneffectof+4wherethesoilfertilityishighand-4whereitislow.

Here,wearrangeblocksaccordingtoFisher’sthreeprinciples.Theblocksarearrangedtoreflectthedifferenceinthesoilfertilitywell.

# set block to consider the heterogeneity of field conditionblock <- c("I", "II", "III", "IV")blomat <- matrix(rep(block, each = 4), nrow = 4)blomat

## [,1] [,2] [,3] [,4]## [1,] "I" "II" "III" "IV"## [2,] "I" "II" "III" "IV"## [3,] "I" "II" "III" "IV"## [4,] "I" "II" "III" "IV"

Next,randomlyarrangevarietiesineachblockaccordingtoFisher’sthreeprinciples.Let’sprepareforthatfirst.

# assume that there are four varietiesvariety <- c("A", "B", "C", "D")

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# sample the order of the four varieties randomlysample(variety)

## [1] "B" "D" "C" "A"

sample(variety)

## [1] "C" "B" "A" "D"

Let’sallocatevarietiesrandomlytoeachblock.

# allocate the varieties randomly to each column of the fieldvarmat <- matrix(c(sample(variety), sample(variety), sample(variety), sample(variety)), nrow = 4)varmat

## [,1] [,2] [,3] [,4]## [1,] "D" "B" "B" "D" ## [2,] "B" "A" "D" "A" ## [3,] "C" "D" "C" "B" ## [4,] "A" "C" "A" "C"

Considerthedifferencesingeneticvaluesofthefourvarieties.LetthegeneticvaluesoftheAtoDvarietiesbe+4,+2,-2,-4,respectively.

# simulate genetic ability of the varietiesg.value <- matrix(NA, 4, 4)g.value[varmat == "A"] <- 4g.value[varmat == "B"] <- 2g.value[varmat == "C"] <- -2g.value[varmat == "D"] <- -4g.value

## [,1] [,2] [,3] [,4]## [1,] -4 2 2 -4## [2,] 2 4 -4 4## [3,] -2 -4 -2 2## [4,] 4 -2 4 -2

Environmentalvariationsaregeneratedasrandomnumbersfromanormaldistributionwithanaverageof0andastandarddeviationof2.5.

# simulate error variance (variation due to the heterogeneity of local environment)e.value <- matrix(rnorm(16, sd = 2.5), 4, 4)e.value

## [,1] [,2] [,3] [,4]## [1,] -1.547611 2.232424 0.1911757 1.8861892## [2,] -2.207789 3.909922 -2.1251164 -0.8860432## [3,] 1.098536 2.524477 -3.2760349 -1.1552031## [4,] 3.110199 -0.690938 -4.2153578 4.7493152

Althoughtheabovecommandgeneratesrandomnumbers,Ithinkyouwillgetthesamevalueasthetextbook.Thisisbecausetherandomnumbersgeneratedarepseudorandomnumbersandaregeneratedaccordingtocertainrules.Notethatifyouchangethevalueoftherandom

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seed,thesamevalueasabovewillnotbegenerated.Also,differentrandomnumbersaregeneratedeachtimeyourun.

Finally,theoverallaverage,thegradientofsoilfertility,thegeneticvaluesofvarieties,andthevariationduetothelocalenvironmentareaddedtogethertogenerateasimulatedobservedvalueofthetrait.

# simulate phenotypic valuesgrand.mean <- 50simyield <- grand.mean + field.cond + g.value + e.valuesimyield

## [,1] [,2] [,3] [,4]## [1,] 48.45239 56.23242 50.19118 43.88619## [2,] 53.79221 59.90992 41.87488 49.11396## [3,] 53.09854 50.52448 42.72397 46.84480## [4,] 61.11020 49.30906 47.78464 48.74932

Beforeperforminganalysisofvariance,reshapedataintheformofmatricesintovectorsandrebundlethem.

# unfold a matrix to a vectoras.vector(simyield)

## [1] 48.45239 53.79221 53.09854 61.11020 56.23242 59.90992 50.52448 49.30906## [9] 50.19118 41.87488 42.72397 47.78464 43.88619 49.11396 46.84480 48.74932

as.vector(varmat)

## [1] "D" "B" "C" "A" "B" "A" "D" "C" "B" "D" "C" "A" "D" "A" "B" "C"

as.vector(blomat)

## [1] "I" "I" "I" "I" "II" "II" "II" "II" "III" "III" "III" "III"## [13] "IV" "IV" "IV" "IV"

Below,thedataisbundledasadataframe.

# create a dataframe for the analysis of variancesimdata <- data.frame(variety = as.vector(varmat), block = as.vector(blomat), yield = as.vector(simyield))head(simdata, 10)

## variety block yield## 1 D I 48.45239## 2 B I 53.79221## 3 C I 53.09854## 4 A I 61.11020## 5 B II 56.23242## 6 A II 59.90992## 7 D II 50.52448## 8 C II 49.30906## 9 B III 50.19118## 10 D III 41.87488

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Let’splotthecreateddatausingthefunctioninteraction.plot.

# draw interaction plotinteraction.plot(simdata$block, simdata$variety, simdata$yield)

Itcanbeseenthatthedifferencebetweenblocksisaslargeasthedifferencebetweenvarieties

Let’sperformananalysisofvarianceusingtheprepareddata.

# perform the analysis of variance (ANOVA) with simulated datares <- aov(yield ~ block + variety, data = simdata)summary(res)

## Df Sum Sq Mean Sq F value Pr(>F) ## block 3 239.11 79.70 11.614 0.00190 **## variety 3 159.52 53.17 7.748 0.00728 **## Residuals 9 61.77 6.86 ## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Itcanbeseenthatboththeblockandvarietyeffectsarehighlysignificant.Notethattheformerisnotthesubjectofverification,andisincorporatedintothemodelinordertoestimatethevarietyeffectcorrectly.

Theanalysisofvariancedescribedabovecanalsobeperformedusingthefunction“lm”forestimatingregressionmodels.

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# perform ANOVA with a linear modelres <- lm(yield ~ block + variety, data = simdata)anova(res)

## Analysis of Variance Table## ## Response: yield## Df Sum Sq Mean Sq F value Pr(>F) ## block 3 239.109 79.703 11.6138 0.001898 **## variety 3 159.518 53.173 7.7479 0.007285 **## Residuals 9 61.765 6.863 ## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Complete random block design and completely randomized design

Thelocalcontrol,oneofFisher’sthreeprinciples,isveryimportantforperforminghighlyaccurateexperimentsinfieldsunderhighheterogeneitybetweenplots.Here,assumingthesameenvironmentalconditionsasbefore,let’sconsiderperforminganexperimentwithoutsettingupablock.

Intheprevioussimulationexperiment,weblockedeachcolumnandplacedA,B,C,Drandomlyinthatblock.Herewewillassigntheplotswith4varietiesx4replicatescompletelyrandomlyacrossthefield.Anexperimentinwhichblocksarenotarrangedintheexplimentandarrangedcompletelyrandomlyiscalled“completelyrandomizeddesign.”

# completely randomized the plots of each variety in a fieldvarmat.crd <- matrix(sample(varmat), nrow = 4)varmat.crd

## [,1] [,2] [,3] [,4]## [1,] "A" "C" "C" "D" ## [2,] "B" "A" "B" "D" ## [3,] "A" "C" "B" "A" ## [4,] "D" "B" "C" "D"

Thistime,youshouldcarefulthatthefrequencyofappearanceofvarietyvariesfromrowtorow,sincevarietiesarerandomlyassignedtotheentirefield.

Thegeneticeffectisassignedaccordingtotheorderofvarietiesinacompletelyrandomarrangement.

# simulate genetic ability of the varietiesg.value.crd <- matrix(NA, 4, 4)g.value.crd[varmat.crd == "A"] <- 4g.value.crd[varmat.crd == "B"] <- 2g.value.crd[varmat.crd == "C"] <- -2g.value.crd[varmat.crd == "D"] <- -4g.value.crd

## [,1] [,2] [,3] [,4]## [1,] 4 -2 -2 -4## [2,] 2 4 2 -4## [3,] 4 -2 2 4## [4,] -4 2 -2 -4

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Asintheprevioussimulationexperiment,theoverallaverage,thegradientofsoilfertility,thegeneticeffectofvarieties,andthevariationduetothelocalenvironmentaresummedup.

# simulate phenotypic valuessimyield.crd <- grand.mean + g.value.crd + field.cond + e.valuesimyield.crd

## [,1] [,2] [,3] [,4]## [1,] 56.45239 52.23242 46.19118 43.88619## [2,] 53.79221 59.90992 47.87488 41.11396## [3,] 59.09854 52.52448 46.72397 48.84480## [4,] 53.11020 53.30906 41.78464 46.74932

Thedataisbundledasadataframe.

# create a dataframe for the analysis of variancesimdata.crd <- data.frame(variety = as.vector(varmat.crd), yield = as.vector(simyield.crd))head(simdata.crd, 10)

## variety yield## 1 A 56.45239## 2 B 53.79221## 3 A 59.09854## 4 D 53.11020## 5 C 52.23242## 6 A 59.90992## 7 C 52.52448## 8 B 53.30906## 9 C 46.19118## 10 B 47.87488

Nowlet’sperformanalysisofvarianceonthedatageneratedinthesimulation.Unlikethepreviousexperiment,wedonotsetblocks.Thus,weperformregressionanalysiswiththemodelthatonlyincludesthevarietaleffectanddoesnotincludetheblockeffect.

# perform ANOVAres <- lm(yield ~ variety, data = simdata.crd)anova(res)

## Analysis of Variance Table## ## Response: yield## Df Sum Sq Mean Sq F value Pr(>F) ## variety 3 218.12 72.705 3.1663 0.06392 .## Residuals 12 275.55 22.962 ## ---## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Intheaboveexample,thevarietaleffectisnotsignificant.Thisisconsideredtobeduetothefactthatthespatialheterogeneityinthefieldcausestheerrortobelargeandthegeneticdifferencebetweenvarietiescannotbeestimatedwithsufficientaccuracy.

Theabovesimulationexperimentwasrepeated100times(shownonthenextpage).Asaresult,intheexperimentusingtherandomcompleteblockdesign,thevarietaleffectwasdetected(thesignificancelevelwassetto5%)in94experimentsoutof100,butitwasdetectedonly66timesinthecompletelyrandomarrangement.Inaddition,whenthe

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significancelevelwassetto1%,thenumberofthevarietaleffectdetectedwas70and30,respectively(inthecaseofcompletelyrandomarrangement,thevarietaleffectwasmissed70times!).Fromthisresult,itcanbeseenthattheadoptionoftherandomcompleteblockdesigniseffectivewhenthereisamong-replicationheterogeneitysuchastheslopeofsoilfertility.Inordertomakeatime-consumingandlabor-intensiveexperimentasefficientaspossible,itisimportanttodesigntheexperimentproperly.

# perform multiple simulationsn.rep <- 100p.rbd <- rep(NA, n.rep)p.crd <- rep(NA, n.rep)for(i in 1:n.rep) { # experiment with randomized block design varmat <- matrix(c(sample(variety), sample(variety), sample(variety), sample(variety)), nrow = 4) g.value <- matrix(NA, 4, 4) g.value[varmat == "A"] <- 4 g.value[varmat == "B"] <- 2 g.value[varmat == "C"] <- -2 g.value[varmat == "D"] <- -4 e.value <- matrix(rnorm(16, sd = 2.5), 4, 4) simyield <- grand.mean + field.cond + g.value + e.value simdata <- data.frame(variety = as.vector(varmat), block = as.vector(blomat), yield = as.vector(simyield)) res <- lm(yield ~ block + variety, data = simdata) p.rbd[i] <- anova(res)$Pr[2] # experiment with completed randomized design varmat.crd <- matrix(sample(varmat), nrow = 4) g.value.crd <- matrix(NA, 4, 4) g.value.crd[varmat.crd == "A"] <- 4 g.value.crd[varmat.crd == "B"] <- 2 g.value.crd[varmat.crd == "C"] <- -2 g.value.crd[varmat.crd == "D"] <- -4 simyield.crd <- grand.mean + g.value.crd + field.cond + e.value simdata.crd <- data.frame(variety = as.vector(varmat.crd), yield = as.vector(simyield.crd)) res <- lm(yield ~ variety, data = simdata.crd) p.crd[i] <- anova(res)$Pr[1]}sum(p.rbd < 0.05) / n.rep

## [1] 0.94

sum(p.crd < 0.05) / n.rep

## [1] 0.54

sum(p.rbd < 0.01) / n.rep

## [1] 0.74

sum(p.crd < 0.01) / n.rep

## [1] 0.21

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Report assignment 1. Usingthenumberofseedsperpanicle(Seed.number.per.panicle)asthedependent

variable𝑦andpaniclelength(Panicle.length)astheindependentvariable𝑥,fitasimpleregressionmodel(𝑦! = 𝜇 + 𝛽𝑥! + 𝜖!)andfindthesampleintercept𝑚andsampleregressioncoefficient𝑏,whichareestimatesof𝜇and𝛽.

2. Testthenullhypothesis𝐻*: 𝛽 = 0.02.3. Testthenullhypothesis𝐻*: 𝜇 = 5.4. Answerthe95%confidenceintervalbetweentheestimatedy,i.e.,𝑦Candthepredictedy,

i.e.,𝑦Xfor𝑥 = 27.5. Fitthepolynomialregressionmodel𝑦! = 𝜇 + 𝛽%𝑥! + 𝛽"𝑥!" + 𝜖!)usingthefirst-and

second-ordertermsof𝑥andanswerthedeterminationcoefficient𝑅"andtheadjustedcoefficientofdetermination𝑅'()" .

6. Comparetheregressionmodelin5withtheregressionmodelin1inananalysisofvarianceandconsiderthevalidityofincludingasecond-ordertermof𝑥intheregressionmodel.

7. Fitthepolynomialregressionmodel𝑦! = 𝜇 + 𝛽%𝑥! + 𝛽"𝑥!" + 𝛽,𝑥!, + 𝜖!)usingthefirst-tothird-ordertermsof𝑥andanswerthedecisioncoefficient𝑅"andtheheadjustedcoefficientofdetermination𝑅'()" .

8. Comparetheregressionmodelin7withtheregressionmodelin5inananalysisofvariancetoexaminethevalidityofextendingthesecond-orderpolynomialregressionmodeltoathird-orderpolynomialregressionmodel.

Submissionmethod:

• CreateareportasapdffileandsubmitittoITC-LMS.• WhenyoucannotsubmityourreporttoITC-LMSwithsomeissues,sendthereportto

report@iu.a.u-tokyo.ac.jp• Makesuretowritetheaffiliation,studentnumber,andnameatthebeginningofthe

report.• ThedeadlineforsubmissionisMay14th.